SOLUTIONS TO MATH 51 MIDTERM 1
January 29, 2004
1.
Find all solutions of the following system:
x
1

x
2
+
x
3
+
2
x
4
=
3
x
2
+
x
3
+
x
4
=
3
x
1
+
x
2
+
3
x
3
+
4
x
4
=
9
Solution.
Write the augmented matrix and then use Gaussian elimination:
1

1
1
2
3
0
1
1
1
3
1
1
3
4
9
→
1

1
1
2
3
0
1
1
1
3
0
2
2
2
6
→
1
0
2
3
6
0
1
1
1
3
0
0
0
0
0
So
x
1
+
2
x
3
+
3
x
4
=
6
x
2
+
x
3
+
x
4
=
3
The free variables are
x
3
and
x
4
, so the solutions are:
x
1
= 6

2
x
3

3
x
4
,
x
2
= 3

x
3

x
4
,
(
x
3
∈
R
, x
4
∈
R
)
2.
Let
L
be the intersection of the two planes
x
+
y
+
z
= 4
and
2
x
+ 3
y
+
z
= 9
.
Find a parametric equation for
L
.
Solution.
Write the augmented matrix and use Gaussian elimination:
fl
fl
fl
fl
1
1
1
4
2
3
1
9
fl
fl
fl
fl
→
fl
fl
fl
fl
1
1
1
4
0
1

1
1
fl
fl
fl
fl
→
fl
fl
fl
fl
1
0
2
3
0
1

1
1
fl
fl
fl
fl
.
so
x
+ 2
z
= 3 and
y

z
= 1, or (moving the free variable
z
to the right hand side)
x
= 3

2
z
and
y
= 1 +
z
. Thus the intersection is given by
x
y
z
=
3

2
z
1 +
z
z
=
3
1
0
+
z

2
1
1
.
3(a)
. Suppose
u
,
v
, and
w
are points in
R
n
such that
k
u
k
=
k
v
k
=
k
w
k
= 1 and
such that
w
=

u
. Suppose also that
v
is not equal to
u
or to
w
. Prove that the
triangle Δ
uvw
has a right angle at
v
.
1
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Solution.
The vector from
u
to
v
is
v

u
.
The vector from
w
to
v
is
v

w
.
We want to show that these two vectors are orthogonal, so we calculate the dot
product:
(
v

u
)
·
(
v

w
) = (
v

u
)
·
(
v
+
u
) =
v
·
v
+
v
·
u

u
·
v

u
·
u
=
k
v
k
2
k
u
k
2
= 1
2

1
2
= 0
3(b)
. Suppose
x
,
y
, and
z
are vectors in
R
n
whose norms are 1, 2, and 3, respec
tively. Suppose each vector is orthogonal (i.e., perpendicular) to each of the other
two. Find a scalar
c
such that the vector
x
+
c
y

z
is orthogonal to the vector
x
+
y
+
z
.
Solution
The vectors
x
+
y
+
z
and
x
+
c
y

z
vectors will be orthogonal provided
their dot product is 0. When we multiply it out (i.e., use the distributive property),
all the “mixed” terms (
x
·
y
,
x
·
z
, etc.) are 0 by orthogonality. So
0 = (
x
+
y
+
z
)
·
(
x
+
c
y

z
) =
x
·
x
+
c
y
·
y

z
·
z
=
k
x
k
2
+
c
k
y
k
2
k
z
k
2
= 1+4
c

9 = 4
c

8
.
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